Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm asked to proved that $x^n-11$ is irreducible in $\mathbb{Q}(\sqrt{-5})[x]$. I have deduced to a point where I have to prove that if $x^n-11$ is reducible in $\mathbb{Q}(\sqrt{-5})[x]$ then it is also reducible in $\mathbb{Z}[\sqrt{-5}][x]$.

Is this fact true? If it is, can someone tell me how to prove that. If not, how else should I approach the original question.

Note that $\mathbb{Z}[\sqrt{-5}][x]$ is not a Unique Factorization Domain! Thanks!

share|cite|improve this question
11 is prime in $\mathbb{Z}(\sqrt{-5})$ – user44322 Dec 3 '12 at 8:16
To invoke Eisenstein, you must first use Gauss's Lemma to work in $\mathbb{Z}[\sqrt{-5}][x]$ but you cant use Gauss's Lemma in non-UFD – user44322 Dec 3 '12 at 8:22
I am completely wrong. Sorry. Since $\mathbb{Z}[\sqrt{-5}]$ is not a UFD, you cannot use irreducibility in $\mathbb{Z}[\sqrt{-5}]$ to show irreducibility in $\mathbb{Q}(\sqrt{-5})$. I deleted my earlier comments. Please see Zach's answer below. It is correct. – Rankeya Dec 3 '12 at 8:22
up vote 4 down vote accepted

(My algebra is not great, so please check this carefully.) We have $[\mathbb{Q}(\sqrt{-5}):\mathbb{Q}] = 2$. Now, $x^n - 11$ is irreducible over $\mathbb{Q}$ by Eisenstein, and so $[\mathbb{Q}(\sqrt[n]{11}):\mathbb{Q}] = n.$ Since $\mathbb{Q}(\sqrt[n]{11})$ is a subfield of $\mathbb{R}$, it follows that $\sqrt{-5}$ is not contained in it. Hence $[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt[n]{11})] = 2.$ But then $[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt{-5})] = n.$ Hence the irreducible polynomial of $\sqrt[n]{11}$ over $\mathbb{Q}(\sqrt{-5})$ must have degree $n$, and so is $x^n-11$. In particular, that polynomial is irreducible.

share|cite|improve this answer
Dear @ZachL.: $[\mathbb{Q}(11^{1/n}):\mathbb{Q}] = n$. – Rankeya Dec 3 '12 at 8:20
Thanks for catching that. The typo appeared in another place, too. – Zach L. Dec 3 '12 at 8:22
$[\mathbb{Q}(\sqrt{-5},\sqrt[n]{11}):\mathbb{Q}(\sqrt{-5})] = n$ Why is this true? – user44322 Dec 3 '12 at 8:32
Nevermind, figured it out. – user44322 Dec 3 '12 at 8:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.